Y = in √ [(2x-1) / (x + 1)], derivation?
y=ln √[(2x-1)/(x+1)]
=(1/2)(ln(2x-1)-ln(x+1))
y'=(1/2)((2/(2x-1))-(1/(x+1))
=3/(2(2x-1)(x+1))
Y = (x ^ 2) * in (Ex), derivation?
y=(x^2)*ln(ex)=x^2+x^2*lnx
y'=2x+2x*lnx+x^2*(1/x)
=3x+2x*lnx
Y = x / 4 ^ x derivation
y'=[x'*4^x-x*(4^x)']/(4^x)^2
=[4^x-x*4^x*ln4]/(4^x)^2
=(1-xln4)/4^x
Y = (the 3rd power of 1-x) / derivative of X under the root sign
Y = (the 3rd power of 1-x) / x under the root sign
=x^(-1/2)-x^(-5/2)
y'=-1/2*x^(-3/2)+5/2*x^(-7/2)
Y = INX / xn derivation Is y equals INX divided by X to the nth power
Y = LNX / x ^ n, then we can know from the derivation rule of quotient that y '= [(LNX)' * x ^ n - LNX * (x ^ n) '] / x ^ 2n and (LNX)' = 1 / x, (x ^ n) '= n * x ^ (n-1), so y' = [1 / X * x ^ n - LNX * n * x ^ (n-1)] / x ^ 2n = 1 / x ^ (n-1) - n * LNX / x ^ (n + 1)
What is INX after derivation
cos2artsinx=1-2sin^2arcsinx=1-2x^2
The derivative result is - 4x. Substitute x = 2 / 2 root sign 2 into the result is - 2 root sign 2